Question

The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its $\frac {1}{16}^{th}$ value?

Answer 1

It is known that,

$t = \frac {2.303}{k} log\ \frac {[R]_0}{[R]}$

$t = \frac {2.303}{60 s^{-1}} log\ \frac {1}{1/16}$

$t = \frac {2.303}{60 s^{-1}} log\ 16$

$t = 4.6 \times 10^{-2} s\ (approximately)$

Hence, the required time is $4.6 × 10^{-2} s$.